Question

Which of the following wavelengths lying in the range $\text{500 nm }$ to $\text{700 nm }$ is in hydrogen spectrum?

• A. $\text{654 nm}$
• B. $\text{730 nm}$
• C. $\text{815 nm}$
• D. $\text{540 nm}$

Answer 1

Answer: (A)

The energy of a photon of wavelength 500 nm is
$\frac{\text{hc}}{\lambda } = \frac{1 2 4 2 \text{eV} \text{nm}}{5 0 0 \text{nm}} = 2 \cdot 4 4 \, \text{eV.}$
The energy of a photon of wavelength 700 nm is
$\frac{\text{hc}}{\lambda } = \frac{1 2 4 2 \text{eV} \text{nm}}{7 0 0 \text{nm}} = 1 \cdot 7 7 \, \text{eV.}$
The energy difference between the states involved in the transition should, therefore, be between 1.77 eV and 2.44 eV.

Figures above diagram shows some of the energies of hydrogen states. It is clear that only those trasitions which end at n = 2 may emit photons of energy between 1.77 eV and 2.44 eV. Out of these only n = 3 → n = 2 falls in the proper range. The energy of the photon emitted in the transition n = 3 to n = 2 is ΔE = (3.4 - 1.5) eV = 1.9 eV. The wavelength is
$\lambda = \frac{\text{hc}}{\Delta \text{E}}$
$= \frac{1 2 4 2 \text{eV} \text{nm}}{1 \cdot 9 \text{eV}}$
= 654 nm.